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We have,

$$\theta = \frac{Kp_\mathrm a}{1+Kp_\mathrm a}$$

where, the fractional coverage $\theta$ is defined as the number of sites occupied by adsorbate $A$ over the total number of sites. $p_\mathrm a$ is the partial pressure of the adsorbate and $K$ is the equilibrium constant. $$\theta = \frac{[A_{\text{ad}}]}{[S_{\text{total}}]}$$

I need to plot the following data to verify the monolayer adsorption of $\ce{CO}$ on 1 g of $\ce{Pt}$ powder.

$$\begin{array}{cc} \hline \text{Equilibrium Pressure (P)/mmHg} & \text{Adsoprtion Amount(V)/ mmHg} \\ \hline 2.50 & 0.507\\ 5.00 & 0.92\\ 10.0 & 1.51\\ 15.0 & 1.97\\ 20.00 & 2.32\\ 30.00 & 2.75\\ 40.00 & 3.13\\ \hline \end{array}$$

The question gives me the following hint

Because adsorption amount $V$ is proportional to $\theta$, $\theta$ can be expressed as $\frac{V}{V_\mathrm m}$ where $V_\mathrm m$ is the adsorption amount at $\theta = 1$

I generated the following graph in Mathematica, simply by plotting $V$ vs $P$

enter image description here

I don't quite understand what the hint wants me to do? Does that require that I normalise the data fashion, i.e by dividing by a particular $V_\mathrm m$, so instead of having $V$ on the y-axis, I have $\theta$ varying from 0 to 1.

Is $V_\mathrm m$ something I obtain from the data/graph by calculating, or is it a tabulated value that I can look up in table somewhere.

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    $\begingroup$ If you plot 1/V vs 1/p, you should get a straight line. How would you interpret the slope and how would you interpret the intercept? $\endgroup$ – Chet Miller Oct 29 '16 at 11:30
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    $\begingroup$ @ChesterMiller I am guessing $V_m$ is the volume of the monolayer? I have generated the appropriate plot, and have attempted to linearise the equation. Is this what you meant? $\endgroup$ – getafix Oct 30 '16 at 6:23
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Self Answer

Additionally, I plotted $\frac{1}{V}$ and $\frac{1}{p}$, as suggested by @Chester Miller in the comments below, and indeed I get a straight line.

enter image description here

A linear model fit was performed and the equation $4.39274 x+0.214246$ fits the data with an $R^2 = 0.999929 $. From my limited knowledge on this topic (I haven't formally studied this subject), the intercept should give me $\frac{1}{\text{adsorption capacity}}$, and the slope should give some indication of the affinity of the adsorption sites?

So going back to the initial equation, $$\theta = \frac{V}{V_\mathrm m} = \frac{Kp_\mathrm a}{1+Kp_\mathrm a} $$

And now considering $$\frac{V_\mathrm m}{V} = \frac{1+Kp_\mathrm a}{Kp_\mathrm a} = \frac{1}{Kp_\mathrm a} + 1$$

and dividing throughout by $V_\mathrm m$ we get,

$$\frac{1}{V} = \frac{1}{V_\mathrm m} + \frac{1}{V_\mathrm mKp_\mathrm a}$$

And we get the appropriate values for $V_\mathrm m$ and $K$ from the slope and intercept of the $1/V$ vs $1/p$ plot, and then obtaining $\theta$ and plotting vs. $p$ is a trivial exercise.

Anyway, using the calculated parameters, I once again plotted $V$ vs $p$ and fit a curve of the form $\frac{V_\mathrm mKp_\mathrm a}{1+Kp_\mathrm a}$

enter image description here

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  • $\begingroup$ I have a question regarding your answer. I have a similar research question, where I want to calculate K and fit the same equation as you did. As a first step, I used your data to test if I can reproduce your results. Unfortunately, I already have problems with your data. The last step (fitting of the curve to the data) does not work with your data in my calculations. Can you please explain in more detail, how you get K from the plot 1/V with 1/p to plot VmKp/1+Kp? $\endgroup$ – Clarissa Mar 11 at 18:49

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